## Scalar curvature equation on warped products manifolds

In this aritcle by Dobarro and Dozo, the authors give another method calculating the scalar curvatrue on warped products manifolds by making using of conformal change.

Let $M = (M^m,g)$ and $N = (N^n,h)$ be two Riemann manifolds. For $\phi\in C^{\infty}(M), \phi> 0$ on $M$, we consider the warped product

$\displaystyle M\times_{\phi} N = ((M\times N)^{m+n}, g + \phi^2 h)$

and show the relationship between the scalar curvatures $R_g$ on $M$, $R_h$ on $N$ and $R_{\phi}$ on $M \times_{\phi} N$. This relationship is a nonlinear partial differential equation satisfied by a power of the weight $\phi$.

Theorem:  Let $R_g, R_h$ and $R_{\phi}$ denote the scalar curvature on $M, N$ and $M \times_{\phi} N$ respectively. Then the following equality holds:

$\displaystyle -\frac{4n}{n+1}\Delta_gu+R_gu+R_hu^{\frac{n-3}{n+1}}=R_{\phi}u\hfill (1)$

where $u=\phi^{\frac{n+1}2}$. Namely,

$\displaystyle R_{\phi}=R_g+\frac{R_h}{\phi^2}-2n\frac{\Delta\phi}{\phi}-n(n-1)\frac{|\nabla_g\phi|^2}{\phi^2}.$

Proof: Write $g_{\phi}\doteqdot g+\phi^2h=\phi^2(\phi^{-2}g+h)$, so $g_{\phi}$ is conformal to $\tilde{g}_{\phi}\doteqdot\phi^{-2}g+h$ on $M\times N$ and $\phi^{-2}g$ is conformal to $g$ on $M$.

Suppose $m \ge 3$, we apply Yamabe equation in $M$ to obtain that $\phi$ satisfies

$\displaystyle -\frac{4(m-1)}{m-2}\Delta_gv+R_gv=v^{\frac{n+2}{n-2}}\widetilde{R}_g\hfill (2)$

with $\phi^2=v^{\frac4{m-2}}$, where $\widetilde{R}_g$ denotes the scalar curvature on $(M^m,\phi^{-2}g)$.
As $m+n\ge3$, we use Yamabe equation in $M\times N$. Hence $\phi$ also satisfies

$\displaystyle -\frac{4(m+n-1)}{m+n-2}\Delta_{\tilde{g}_{\phi}}w+\widetilde{R}_{g_{\phi}}w={R}_{g_{\phi}}w^{\frac{m+n+2}{m+n-2}}\hfill (3)$

with $w^{\frac4{m+n-2}}=\phi^2$ where $\widetilde{R}_{g_{\phi}}$ denotes the scalar curvature on $(M^m+N^n,\tilde{g}_{\phi})$ and $\Delta_{\tilde{g}_{\phi}}$ the corresponding laplacian.

From $w\in C^{\infty}(M)$, we deduce that

$\displaystyle \Delta_{\tilde{g}_{\phi}}w=\Delta_{\phi^{-2}g+h}w=\Delta_{\phi^{-2}g}w.$

Working in local coordinates

$\displaystyle \Delta_{\phi^{-2}g}w=\frac1{\sqrt{|\phi^{-2}g|}}\partial_i\left((\phi^{-2}g)^{ij}\sqrt{|\phi^{-2}g|}\partial_jw\right)$

where $|\phi^{-2}g|=\det(\phi^{-2}g_{ij})=\phi^{-2m}|g|$ and $(\phi^{-2}g)^{ij}=\phi^2g^{ij}$, hence

$\displaystyle\begin{gathered} \Delta_{\phi^{-2}g}w=\phi^m\frac1{\sqrt{|g|}}\partial_i\left(\phi^{2-m}g^{ij}\sqrt{|g|}\partial_jw\right)\hfill\\ \qquad=v^{\frac{-2m}{m-2}}\frac1{\sqrt{|g|}}\partial_i\left(v^2g^{ij}\sqrt{|g|}\partial_jw\right)\hfill\\ \qquad=\left(v\Delta_gw+2g^{ij}\partial_iv\partial_jw\right)v^{-\frac{n+2}{n-2}}. \end{gathered}$

On the other hand,

$\displaystyle \Delta_g(vw)=v\Delta_gw+w\Delta_gv+2g^{ij}\partial_iv\partial_jw.$

Hence

$\displaystyle v^{\frac{n+2}{n-2}}\Delta_{\phi^{-2}g}w=\Delta_g(vw)-w\Delta_gv.\hfill(4)$

We have

$\displaystyle \widetilde{R}_{g_{\phi}}=\widetilde{R}_g+R_h\hfill(5)$

By using of this in (3)  and multiplying by $v^{\frac{n+2}{n-2}}$, we obtain from (4)

$\displaystyle -\frac{4(m+n-1)}{m+n-2}\big(\Delta_g(vw)-w\Delta_gv\big)+v^{\frac{n+2}{n-2}}(\widetilde{R}_g+R_h)={R}_{g_{\phi}}w^{\frac{m+n+2}{m+n-2}}v^{\frac{n+2}{n-2}}\hfill (6)$

From (2) we arrive at

$\displaystyle\begin{gathered} -\frac{4(m+n-1)}{m+n-2}\Delta_g(vw)-\frac{4n}{(m+n-2)(m-2)}w\Delta_gv\hfill\\ \qquad+R_g(vw)+R_hwv^{\frac{n+2}{n-2}}=R_{\phi}(vw)w^{\frac4{m+n-2}}v^{\frac4{m-4}}=R_{\phi}(vw). \end{gathered}$

Recalling that $w^{\frac4{m+n-2}}v^{\frac4{m-4}}=\phi^2\phi^{-2}=1$ and denoting $u =\phi^{\frac{n+1}2}$, replacing in terms of $u$ in this last equality, then multiplying by $u^{1/n+1}$, we obtain

$\displaystyle\begin{gathered} -\frac{4(m+n-1)}{m+n-2}\Delta_g(u^{\frac n{n+1}})u^{\frac1{n+1}}-\frac{4n}{(m+n-2)(m-2)}u^{\frac{m+n-1}{n+1}}\Delta_gu^{\frac{2-m}{n+1}}\hfill(7)\\ \quad+R_gu+R_hu^{\frac{n-3}{n+1}}=R_{\phi}u. \end{gathered}$

For any $\alpha\neq0$, $u\in C^{\infty}(M)$, $u>0$ such that

$\displaystyle \Delta_gu^{\alpha}=\alpha(\alpha-1)u^{\alpha-2}|\nabla_gu|^2+\alpha u^{\alpha-1}\Delta_gu.$

Choosing $\alpha=\frac n{n+1}$ and $\frac{2-m}{n+1}$ in (7), we obtain

$\displaystyle -\frac{4n}{n+1}\Delta_gu+R_gu+R_hu^{\frac{n-3}{n+1}}=R_{\phi}u.$